STRUCTURE OF Nb-93
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. STRUCTURE OF Nb-93 Naturally occurring niobium (Nb), element 41, is composed of one stable isotope (93Nb). In general Nb with 41 protons (odd number of protons) brakes the high symmetry of the structure of Zr with 40 protons. So it is composed of only one stable nuclide. Following the diagram of Zr-94 with S =0 you see that the 36 deuterons from p1n1 to p36n36 have the structure of a core representing a complete shape of six horizontal planes . Looking at the diagram of the top view of the first plane one concludes that each plane has 6 deuterons and 2 extra neutrons (n) of weak bonds. That is, the core of 36 deuterons has 12 extra neutrons (n) of weak bonds. Note that all nucleons of the 6 planes have spin S= 0 because the first plane is characterized by positive spins, while the sixth plane has nucleons of negative spin. For revealing the structure of Nb-93 with S= +9/2 one sees that the additional deuteron p41n41 cannot give any stable structure when it fills the blank positions in front of p38n38 or behind the n40p40. In other words the addition of p41n41 in front of the square or behind it produces a new fundamental rearrangement of nucleons in order to have the proper symmetry able to give the stability of Nb. Under this condition the deuterons p37n37 and p39n39 change their spins from S =-2 to S=+2 in order to fill the symmetrical blank positions in front of p38n38 and behind the p40n40. Then adding the p41n41 of S =+1 at the same plane we get the total spin S = +4. Therefore taking into account that in this structure the number of extra neutrons is 11, one can add 7 extra neutrons of positive spin and 4 extra neutrons of negative spin to get the total spin of Nb-93 as S = + 4 + 7(+1/2) +4(-1/2) = +9/2. ' DIAGRAM OF Zr WITH 40 DEUTERONS , 8 EXTRA n AND 12 EXTRA (n)' Here the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Also the 12 extra neutrons (n) with weak bonds of six planes and the 4 extra neutrons n of strong bonds near the p37 and p38 are not shown. You can see only 4 extra neutrons n of strong bonds existing under the p21 and p22 and over the p31 and p32. ' ' n40.........p40.......n ' n........p38..........n38 Horizontal squarewith n ' ' n31………p12..........n12........p32' ' p31....... n11.........p11…… n32 Sixth horizontal plane' ' p29....... n10.........p10…….... n30' ' n29………..p9..........n9 …….p30 Fifth horizontal plane' ' n27.........p8..........n8...........p28' ' p27.........n7..........p7........n28 Fourth horizontal plane' ' p25.........n6.........p6..........n26' ' n25……….p5........n5……….p26 Third horizontal plane' ' n23………p4........n4………….p24' ' p23……..n3………p3………..n24 Second horizontal plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First horizontal plane' ' n'.........p37 ......n37 ' ' n39......p39.........n Horizontal square with n ' ' ' '''TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ''' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' ' ' Category:Fundamental physics concepts